If \(f\) is a continuous function and if \(F'(x)=f(x)\) for all real numbers \(x\), then \(\displaystyle \int\limits_1^2 f(3x)   dx =\)

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Integrate using \(u\) substitution.

$$ u=3x \hskip{2em} du = 3 dx \hskip{2em} dx = \frac{1}{3}du $$ $$ \int\limits_1^2 f(3x)   dx = \int\limits_3^6 f(u)   \frac{1}{3}du $$

We are given that \(F'(x) =f(x)\). Integration of both sides implies \( F(x)=\int f(x)\).

$$ \frac{1}{3}\int\limits_3^6 f(u)   du= \frac{1}{3}\Big[F(u)\Big]_3^6 $$ $$ = \boxed{\frac{1}{3}F(6)-\frac{1}{3}F(3)} $$